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CHAPTER 3 – INTRODUCTION TO ENTROPY Instructor: Prof. Dr. Uğur Atikol
CHAPTER 3 – INTRODUCTION TO ENTROPY
Instructor:
Prof. Dr. Uğur Atikol
Definition of Entropy
Entropy is a measure of disorder of a system
Entropy is created during a process
Entropy can not be destroyed
The Clausius Inequality
This inequality is valid for all cycles, reversible or irreversible
Cyclic integral = for reversible
< for irreversible
The cyclic integral of δQ/T can be viewed as
the sum of all the differential amounts of heat
transfer divided by the temperature at the
boundary.
𝛿𝑄
𝑇≤ 0
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
For reversible cycles:
0)(:
0
11
)(
rev
L
L
H
H
L
L
H
H
L
L
H
Hrev
H
H
L
L
H
L
H
L
T
Qnote
T
Q
T
Q
QT
QT
T
Q
T
Q
T
Q
T
Q
T
Q
T
T
Q
Q
Rev.H. E.
Wnet
QL
QH
High temp. reservoir at TH
High temp. reservoir at TL
Reversible Cycles
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
For irreversible cycles:
LirrevL QQ , or DiffLirrevL QQQ ,
0)(
0
)(,
irrev
L
Diff
L
Diff
L
L
H
H
L
irrevL
H
Hirrev
T
Q
T
Q
T
Q
T
Q
T
Q
T
Q
T
Q
T
Q
For all cycles, the two results are combined:
Note: violates the 2nd law of thermodynamics
has to be always negative.
QDiff
Rev.H. E.
IRREV.
H. E.
Wnet Wnet, irrev
QL, irrevQL
QHQH
High temp. reservoir at TH
High temp. reservoir at TL
𝛿𝑄
𝑇≤ 0
𝛿𝑄
𝑇>0
𝛿𝑄
𝑇
0dVInternally reversible
The net change in volume (a property) during a cycle is
always zero.
Any property change during a cycle is zero.
Since must represent a
property in the differential form.
revintrevint
,0
T
Q
T
Q
BA
BA
T
Q
T
Q
T
Q
T
Q
T
Q
)()(
0)()()(
2
1
2
1
1
2
2
1
revint
Entropy is a property
)K
kJ()( revintT
QdS
Entropy is an extensive property.
The entropy change of a system: )K
kJ()( revint
2
1
12 T
QSSS
Example: air temperature is raised from T1 to T2
Air
1
2
2
1
revint
2
1
ln)(T
TCm
T
dTCm
T
QS
dTCmUdQ
UdWQ
vv
v
0
Q
Thermal
insulation
This property is called entropy, S .
The value of the integral depends on the end states only and not the path followed
This represents the change of a property
Liq+vap mixture
T = 300K = const.
Q = 750 kJ
KkJ
)(1
)()(
2
1
revintrevint
2
1
revint
2
1
C
CC
T
QS
QTT
Q
T
QS
Constant absolute temperature
Particularly useful for determining the entropy changes of thermal energy reservoirs that can absorb or supply heat indefinitely at constant temperature.
kJ/K 2.5
T
QSsys
Special Case: Internally Reversible Isothermal heat transfer processes:
The inequality dS Q/T implies that for irreversible cases dS is greater than Q/T
Therefore dS Q/T > 0 and this quantity is known as entropy generation
For any closed system:
Increase of Entropy Principle: Entropy Generation
gen
n
i i
isys
gensys
ST
QdSd
ST
QdSd
1
:boundary on the positions
fer heat trans several are thereifor
Tn T2 T1
Q1 Q2 Qn
Closed system
Increase of Entropy Principle: Entropy Generation Consider equation
For an isolated system SQ=0 0
The entropy of an isolated system always increases (due to irreversibilities) or if reversible, remains constant.
Entropy Balance
The property entropy is a measure of molecular disorder or randomness of a system.
Enropy can be created but it cannot be destroyed
systemgenoutin SSSS
or
system theof
entropy total
in the Change
generated
entropy
Total
leaving
entropy
Total
entering
entropy
Total
Entropy Change of a System Ssys
dVsmsS
S
SSS
sys
sys
initialfinalsys
:by determined becan system theof
entropy theuniform,not are system theof properties When the
operation. statesteady during 0 :Note
state initialat Entropy state finalat Entropy system a of changeEntropy
density
volume
Entropy Change of Pure Substances TdS relations:
ℎ = 𝑢 + 𝑃𝑣 → 𝑑ℎ = 𝑑𝑢 + 𝑃𝑑𝑣 + 𝑣𝑑𝑃𝑇𝑑𝑆 = 𝛿𝑄 → 𝑇𝑑𝑠 = 𝑑𝑢 + 𝑃𝑑𝑣
𝑇𝑑𝑠 = 𝑑ℎ − 𝑣𝑑𝑃
Hence useful relations can be obtained for ds:
𝑑𝑠 =𝑑𝑢
𝑇+
𝑃𝑑𝑣
𝑇 and 𝑑𝑠 =
𝑑ℎ
𝑇−
𝑣𝑑𝑃
𝑇
We must know the relationship between du or dh and T
𝑑𝑢 = 𝑐𝑣𝑑𝑇 or for ideal gases 𝑑ℎ = 𝑐𝑝𝑑𝑇
For ideal gases: 𝑃𝑣 = 𝑅𝑇 and hence:
𝑑𝑠 = 𝑐𝑣𝑑𝑇
𝑇+ 𝑅
𝑑𝑣
𝑣 𝑜𝑟 𝑑𝑠 = 𝑐𝑝
𝑑𝑇
𝑇− 𝑅
𝑑𝑃
𝑃
For liquids and solids assume incompressible hence 𝑑𝑣 ≅ 0
Hence for liquids and solids: 𝑑𝑠 =𝑑𝑢
𝑇=
𝑐𝑑𝑇
𝑇 since 𝑐𝑝 = 𝑐𝑣 = 𝑐
and 𝑑𝑢 = 𝑐 𝑑𝑇
Entropy Change of Ideal Gases Specific heats vary with temperature
Therefore
𝑑𝑠 = 𝑐𝑝𝑑𝑇
𝑇− 𝑅
𝑑𝑃
𝑃→ 𝑠2 − 𝑠1 = 𝑐𝑝(𝑇)
𝑑𝑇
𝑇− 𝑅
𝑑𝑃
𝑃
2
1
Choose absolute zero as reference T and define:
𝑠° = 𝑐𝑝(𝑇)𝑑𝑇
𝑇
𝑇
0
Table A17 in Çengel* tabulate 𝑠°
Therefore 𝑐𝑝 𝑇𝑑𝑇
𝑇=
2
1𝑠°2 − 𝑠°1
Hence 𝑠2 − 𝑠1 = 𝑠°2 − 𝑠°1 − 𝑅𝑑𝑃
𝑃
Mechanisms of entropy transfer, Sin and Sout Entropy can be transferred by the following two
mechanisms:
Heat transfer
Mass flow
No entropy is transferred by work
Heat is a chaotic form of energy and some chaos
(entropy) flows with heat
Mass contains entropy and entropy is carried with it.
Entropy increases with mass
Entropy transfer by heat transfer
Entropy transfer by work:
Heat transfer is always accompanied by entropy transfer in the amount of Q/T, where T is the boundary temperature.
No entropy accompanies work as it crosses the system boundary. But entropy may be generated within the system as work is dissipated into a less useful form of energy.
When temperature is not constant or
different throughout the boundary
Entropy transfer by mass flow
Entropy transfer by mass:
Mass contains entropy as well as energy, and thus mass flow into or out of system is always accompanied by energy and entropy transfer.
When the properties of the mass change during the process
Entropy generation, Sgen
System Ssys
Sgen≥0
Sin with mass Sin with heat
Sout with mass Sout with heat
The term Sgen represents the entropy within the system boundary only
External irreversibilities are not accounted for in the term Sgen.
Entropy generation, Sgen
Entropy generation outside system
boundaries can be accounted for by
writing an entropy balance on an
extended system that includes the system
and its immediate surroundings.
Entropy balance of control masses (closed systems)
(kJ/K)
generatedEntropy
ferheat transbyboundary system
hrough thetransfer tentropynet of Sumsystem closed a of
changeEntropy
gen
k
ksys S
T
QS
System
T1 T2 T3
T0
W
0Q
1Q2Q 3Q
Example: Entropy generation in a wall
W/K191.0 therefore 0K 278
W1035
K 293
W1035
0
,
entropyin change
of Rategenerationentropy of Rate
mass andheat by transfer
entropynet of Rate
wallgengen
gen
outin
sys
genoutin
SS
ST
Q
T
Q
dt
SdSSS
Determine the rate of entropy generation in a wall of 5-m x 7-m and thickness 30 cm. The rate of heat transfer through the wall is 1035 W.
The total rate of entropy generation (including the indoors and outdoors) can be found by taking into account the indoors and outdoors temperatures (extended system):
W/K341.0 therefore 0K 273
W1035
K 300
W1035 gen, totalgen SS
0 (steady heat flow)
Entropy balance of control volumes (open systems)
(kW/K)
:form rate in theor
(kJ/K) )(
volumecontrol theinon accumulati
entropy of Rate
CV
rategenerationEntropy
flow mass via volumecontrol theofout rate flowentropy Net
transferheat by rate
ansfer Entropy tr
CV12
CV
dt
dSSsmsm
T
Q
SSSsmsmT
Q
geneeii
k
k
S
geneeii
k
k
The entropy of a control volume changes as a result of mass flow as well as heat transfer.
inmoutmSystem
T1 T2 T3
T0
W
0Q
1Q2Q 3Q
Entropy balance of control volumes (open systems)
The entropy of a control volume changes as a result of mass flow as well as heat transfer.
CV
dt
dSSsmsm
T
Qgeneeii
k
k
The entropy of a substance always increases (or remains constant in the case of a reversible process) as it flows through a single-stream, adiabatic, steady-flow device.
Example: Entropy generation during a throttling process Determine the rate of entropy generation in a steady-state throttling process of steam shown in the diagram. Use the tables to determine the entropy at the inlet and the exit states:
kJ/kg.K 0046.7 MPa 3
:2 State
kJ/kg.K 6353.6 kJ/kg, 3.3288 C450
MPa 7 :1 State
2
12
2
11
1
1
shh
P
shT
P
volumecontrol in theentropy in
change of Rategenerationentropy of Rate
flowmassby transfer
entropy of Rate
CV
dt
SdSSS
dt
dSSsmsm
T
Q
sys
genoutin
geneeii
k
k
kJ/kg.K 3693.06353.60046.7
:rate flow massby Dividing
)(
0
12
12
21
sss
ssmS
Ssmsm
gen
gen
gen
0 (negligible heat transfer) 0 (steady flow process)
Example: Entropy generation in a compressor
Compressor300 kW
25 kW
Air 𝑚 1 = 0.853 kg/s 𝑃1 = 100 kPa 𝑇1 = 𝑇𝑎𝑚𝑏 = 17℃
𝑃1 = 1 MPa 𝑇1 = 327℃
kW/K 155.0K 290
kW 25kW/K 0684.0
kW/K 0684.0
kPa 100
kPa 1000ln 287.0
kg.K
kJ)66802.1(2.40902 kg/s 853.0)(
ln :gases idealFor
)(
0
12
1
20
1
0
212
,
12
,
21
entropyin change
of Rategenerationentropy of Rate
mass andheat by transfer
entropynet of Rate
gen
air
surrb
outgen
gen
surrb
out
sys
genoutin
S
ssm
P
PRssss
T
QssmS
ST
Qsmsm
dt
SdSSS
0 (steady flow process)
kJ/kg.K 40902.22 os
kJ/kg.K 66802.11 os
Example: Entropy transfer associated with heat transfer A frictionless piston-cylinder contains saturated liquid vapor mixture at 100oC. 600kJ is lost to the environment at constant pressure leading to condensation of some vapor.
kJ/K 61.1273)K(100
kJ 600
: waterof changeentropy The
sys
sysT
QS
kJ/K 0.40kJ/K) 61.1(K )27325(
kJ 600
:changeentropy totalfor system extended thegConsiderin
system theofchangeEntropy
generationEntropy
mass andheat by ansferentropy trNet
sys
b
outgensysgen
b
out
sysgenoutin
ST
QSSS
T
Q
SSSS
The extended system includes the water, the piston-cylinder device and the surroundings just outside the system that undergoes a temperature change. The boundary of the extended system is at Tsurr.
Entropy generation associated with a heat transfer process Pinpointing the location of entropy generation: Be more precise about the system, the b0undary and the surroundings.
Homework Steam expands in a turbine steadily at a rate of 40,000
kg/h, entering at 8 MPa and 500oC and leaving at 40 kPa as saturated vapor. If the power generated by the turbine is 8.2 MW, determine the rate of entropy generation for this process. Assume the surrounding medium is at 25oC.
8.2 MW
8 MPa 500oC
40 kPa Sat. vapor
Steam turbine
